Name: Linear Momentum of a System of Particles
Uploaded: 2019-07-19
Duration: 10 min 32 s
Description: Linear momentum is one of the most important concepts in physics. Watch this video to dive deep to visualise linear momentum of a system of particles.

Question

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of spring constant $k$ . Suppose the block of mass $m_{1}$ is pulled by a constant force $F_{1}$ and the other block is pulled by a constant force $F_{2}$ . Find the maximum elongation that the spring will suffer.

Accepted Answer

Given: The block of mass $m_{1}$ is pulled by a constant force $F_{1}$ and other block of mass $m_{2}$ is pulled by a constant force $F_{2}$ .

Let $F_{2} > F_{1}$ .

Then, acceleration of the centre of mass, $a_{c} = \frac{F_{2} - F_{1}}{m_{1} + m_{2}}$

At maximum elongation, velocities of both the blocks become equal. Therefore, velocities of both becomes equal to the velocity of the centre of mass at that instant.

So, both the blocks will be at rest with respect to the centre of mass at maximum elongation.

Therefore, to calculate the maximum elongation, consider the system with respect to the centre of mass.

In the centre of mass frame, along with the real forces, pseudo forces also act on both the blocks, as shown in the figure.

Net force on the block of mass $m_{1}$ , $F_{1 N} = F_{1} + m_{1} a_{c} = F_{1} + m_{1} (\frac{F_{2} - F_{1}}{m_{1} + m_{2}})$

$∴ F_{1 N} = \frac{m_{1} F_{2} + m_{2} F_{1}}{m_{1} + m_{2}} . . . (1)$

Net force on the block of mass $m_{2}$ , $F_{2 N} = F_{2} - m_{2} a_{c} = F_{2} - m_{2} (\frac{F_{2} - F_{1}}{m_{1} + m_{2}})$

$∴ F_{2 N} = \frac{m_{1} F_{2} + m_{2} F_{1}}{m_{1} + m_{2}} . . . (2)$

Let the maximum elongation in the left part and right part of the spring be $x_{1}$ and $x_{2}$ , respectively.

Applying work energy theorem, the maximum elongation in the spring is,

$\frac{1}{2} k {(x_{1} + x_{2})}^{2} = (F_{1 N} \times x_{1}) + (F_{2 N} \times x_{2})$

From equations $(1)$ and $(2)$ , $F_{1 N} = F_{2 N}$

$∴$ $x_{1} + x_{2} = \frac{2 F_{1 N}}{k}$

Therefore, maximum elongation in the spring, $x = x_{1} + x_{2} = \frac{2 (m_{1} F_{2} + m_{2} F_{1})}{(m_{1} + m_{2}) k}$

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of spring constant $k$ . Suppose the block of mass $m_{1}$ is pulled by a constant force $F_{1}$ and the other block is pulled by a constant force $F_{2}$ . Find the maximum elongation that the spring will suffer.

Important Questions on Centre of Mass, Linear Momentum, Collision

Learn and Prepare for any exam you want !

Two blocks of masses m1 and m2 are connected by a spring of spring constant k. Suppose the block of mass m1 is pulled by a constant force F1 and the other block is pulled by a constant force F2. Find the maximum elongation that the spring will suffer.

Important Questions on Centre of Mass, Linear Momentum, Collision

Learn and Prepare for any exam you want !

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of spring constant $k$ . Suppose the block of mass $m_{1}$ is pulled by a constant force $F_{1}$ and the other block is pulled by a constant force $F_{2}$ . Find the maximum elongation that the spring will suffer.